Answer
$$\int \sin3x\cos2xdx=-\frac{\cos5x}{10}-\frac{\cos x}{2}+C$$
Work Step by Step
$$A=\int \sin3x\cos2xdx$$
Here we need to use the trigonometric identity: $$\sin a\cos b=\frac{\sin(a+b)+\sin(a-b)}{2}$$
Therefore, $$A=\int\frac{\sin5x+\sin x}{2}dx$$ $$A=\frac{1}{2}\Big(\int\sin5xdx+\int\sin xdx\Big)$$ $$A=\frac{1}{2}\Big(-\frac{\cos5x}{5}-\cos x\Big)+C$$ $$A=-\frac{\cos5x}{10}-\frac{\cos x}{2}+C$$
